Your solution for the roots of the homogeneous equation was correct and leads to
y = Ae(-1+i√5)t+Be(-1-i√5)t
You can rewrite that as
y = e-t(C cos(αt) + D sin(αt)) where α=√5.
Don't confuse that α with the given ω.
Now we just have to find a particular solution for the inhomogeneous equation. Clearly this will be of the form A cos(ωt) + B sin(ωt) (different A and B from before).
you correctly obtained
-ω2A-2ωB+6A=10
-ω2B+2ωA+6B=0
but I don't understand where you went from there. How did you deduce B = 0?
Wrt q 2, what is γ in this context?

Your solution for the roots of the homogeneous equation was correct and leads to
y = Ae(-1+i√5)t+Be(-1-i√5)t
You can rewrite that as
y = e-t(C cos(αt) + D sin(αt)) where α=√5.
Don't confuse that α with the given ω.
Now we just have to find a particular solution for the inhomogeneous equation. Clearly this will be of the form A cos(ωt) + B sin(ωt) (different A and B from before).
you correctly obtained
-ω2A-2ωB+6A=10
-ω2B+2ωA+6B=0
but I don't understand where you went from there. How did you deduce B = 0?
Wrt q 2, what is γ in this context?

-ω2A-2ωB+6A=10
-ω2B+2ωA+6B=0 I can't solve it.I had a mistakes.I just figure now.There's no γ in my context.